Question

In a class of 200 students who appeared certain examinations, 35 students failed in MHT-CET, 40 in AIEEE and 40 in IIT entrance, 20 failed in MHT-CET and AIEEE, 17 in AIEEE and ITT entrance, 15 in MHT-CET and IIT entrance and 5 failed in all three examinations. Find how many students
i) did not fail in any examination.
ii) failed in AIEEE or IIT entrance.

Answer 1

Answer:

i) 132

ii) 63

Step-by-step explanation:

Hi,

Let n(M) be the number of students who failed in MHT-CET

n(M) = 35

Let n(A) be the number of students who failed in AIEEE,

n(A) = 40

Let n(I) be the number of students who failed in IIT,

n(I) = 40

Given  20 failed in MHT-CET and AIEEE,

n( M ∩ A) = 20

Given  17 failed in AIEEE and IIT entrance,

n(A ∩ I) = 17,

Given 15 failed in MHT-CET and IIT entrance

n(M ∩ I) = 15

Given 5 failed in all three examinations,

n(A ∩ M ∩ I) = 5

i) To find number of student who did not fail in any of the examination

= [n(A ∪ M ∪ I)']

= n(x) - n(A ∪ M ∪ I)

n(A ∪ M ∪ I) = n(A) + n(M) + n(I) - n(A∩M) - n(A∩I) - n(M∩I) + n(A∩M∩I)

= 35 + 40 + 40 - 20 - 17 - 15 + 5

= 68

So n(A∪M∪I) = 68

Number of students who did not fail in any examination will be

200 - 68

= 132

ii) Students failed in AIEE or IIT

= n(A∪I)

= n(A) + n(I) - n(A∩I)

= 40 + 40 - 17

= 63

Hope, it helps !

Answer 2

Answer:

132 are not failed 63 are failed in